$R\subseteq S$ integral extension and $S$ Noetherian implies $R$ Noetherian?

Question

The problem is as follows:

Let $R\subseteq S$ be an integral extension and $S$ a Noetherian ring.

1. Then show that for each $\mathfrak p\in \operatorname{Spec}R$, there are only finitely many $P\in \operatorname{Spec}S$ such that $P\cap R = \mathfrak p$.

2. Is $R$ also Noetherian?

I am able to show the first part, but unable to show the second part. Any help is welcome.

Answer 1

2. Consider the field $$ \widetilde{\mathbb{Q}}=\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5},\ldots)=\mathbb{Q}(\sqrt{p}\mid\;p\mbox{ is positive prime integer}). $$ The rings $R=\mathbb{Q}+X\widetilde{\mathbb{Q}}[X]$ and $S=\widetilde{\mathbb{Q}}[X]$ provide a counterexample to your question.